Why does subtracting an eigenvalue from the diagonal and solving give the eigenspace?

Question

Given a matrix $A$, with eigenvalue $\lambda$, why does substracting $\lambda I$ from $A$, and then solving the matrix give us the eigenspace?

Like, I don't understand the intuition behind subtracting the two, to hence obtain the eigenspace.

I can't find an answer anywhere, a high level explanation would be very much appreciated! Thank you in advance.

Answer 1

Let $V$ a vector space. The eigenspace associated to the eigenvalue $\lambda $ is the set $E_\lambda $ of vector $v\in V$ s.t. $$Av=\lambda v.$$ In otherword, $$E_\lambda =\{v\in V\mid Av=\lambda v\}.$$

Therefore $$v\in E_\lambda \iff Av=\lambda v\iff (A-\lambda I)v=Av-\lambda v=0$$ $$\iff v\in \ker(A-\lambda I).$$

We conclude that $$E_\lambda =\ker(A-\lambda I).$$

Answer 2

Let $V$ be a $n$-dimensional vector space over any field and $A$ any $(n \times n)$-Matrix acting on that space. Remember the eigenvalue equation $$Av = \lambda v$$ for $v \neq 0$. Now, rewrite $v=I v$, where $I$ is the identity matrix, and substract it from the eigenvalue equation, hence $$(A- \lambda I )v = 0.$$ Let us assume for a moment that the matrix $A-\lambda I$ is invertible. For that we'd multiply both sides with the inverse and we'd get $v=0$. A contradiction to our requirement that $v \neq 0$. Therefore, $A-\lambda I$ cannot be invertible and so $\det(A-\lambda I) = 0$. But the characteristic polyomial is exactly the determinant of $A-\lambda I$.

On the other hand, if $\lambda$ is a zero of the characteristic polynomial, i.e. $A-\lambda I$ is not invertible, $(A-\lambda I) v =0$ has a non-trivial solution and its solution is an eigenvector corresponding to the eigenvalue $\lambda$.

Answer 3

Notice that \begin{align*} Av= \lambda v,~v \ne 0 \Longleftrightarrow (A - \lambda I)v = 0,~v \ne 0 \Longleftrightarrow v \in\text{Nul}(A- \lambda I),~v \ne 0. \end{align*}